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1 year ago

The area of a quadrilateral whose vertices is ABCD taken in order are (1,2) (-5,6) (7,-4) and (-2,t) be 0 ,find the value of t

1 answer:

3 0

Coordinates : A(1,2) B(-5,6) C(7,-4) D(-2,t) - All are taken in order

gradient of AB = (6-2)/(-5-1) = 4/-6 = -⅔
since it is quadrilateral, AB//CD
gradient of CD = -⅔
equation of CD : y-(-4)=-⅔(x-7)
y+4=-⅔x+14/3
y=-⅔x+⅔
now, sub x=-2 from D(-2,t)
since t is the y-coordinate of point D
t=-⅔(-2)+⅔
t=2

Therefore, t=2

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